The mean and standard deviation of a binomial variate X are 4 and `sqrt(3)` respectively. Then ,`P(Xge1)=`

Options

A. `1-((1)/(4))^(16)`
B. `1-((3)/(4))^(16)`
C. `1-((2)/(3))^(16)`
D. `1-((1)/(3))^(16)`

Answer

B

Explanation

(b) We have, mean, np=4 and standard deviation `=sqrt(3)`.
So, variance, npq=3
`:. " " q=(3)/(4), p=(1)/(4)` and n=16
Now, `P(X ge 1)=1-(X=0)=1""^(16)C_(0)((3)/(4))^(16)=1-((3)/(4))^(16)`